Question

A survey reveals that 25 percent of a population of 1,000 individuals have attached earlobes (are homozygous recessive for the trait). For the following questions, assume that the population fits the parameters of the Hardy-Weinburg law.

If p equals the frequency of the dominant allele and q equals the frequency of the recessive allele, which of the following terms represents the frequency of the individuals who show the dominant phenotype?
a. pxp
b. qxq
c. 2pq
d. pxp +2pq
e. pxp + qxq

Answer 1

d. pxp +2pq

Step-by-step explanation:

The formula for genotype frequency for a population in Hardy-Weinburg equilibrium is as under:

p² + 2pq + q² = 1

where, p = dominant allele

q = recessive allele

Here,

p² represents frequency of homozygous dominant genotype

2pq represents frequency of heterozygous genotype

q² represents frequency of homozygous recessive genotype

Also, although the genotypes p² & 2pq are different from each other yet phenotypically they both will collectively produce dominant trait i.e. free ear lobes not attached earlobes. So the term "p² + 2pq or pxp + 2pq" represents the frequency of the individuals who show the dominant phenotype in this particular population. Dominant phenotype will comprise 75% of the population.